PH8151 Engineering Physics Chapter 1

1.9. BENDING MOMENT OF A BEAM

OP=OQ=R and \POQ=◊such thatPQ=R◊

Consider another filament DE at a distancexabove MN and two pointsPÕandQÕon DE
as shown in Figure1.24. Then,

PÕQÕ=(R+x)◊

The strain at the layer DE =

change in length

original length

=

PÕQÕ≠PQ

PQ

So, linear strain =

[(R+x)◊]≠(R◊)

R◊

=

x◊

R◊

=

x

R

(1.26)

Since, stress = (YoungÕs modulus(Y)◊strain), we have from equation (1.26),

stress =Y

(^3) x
R
4
(1.27)
Hence, the force acting on the elemental area”Aof the cross-section of the beam which
is at a distancexfrom the neutral axisNNÕis
stress ◊ area =
Yx
R
◊”A
Moment of the force”Fabout the neutral axisNNÕis
(Force)◊(‹ distance from axis to line of action of force) =
Yx”A
R
◊x=
Y
R
(”Ax^2 )
The cross-sectional area of the beam can be thought of as made up of a large number
of small elemental rectangles each of area”A. Therefore, the total moment of the forces
acting on the cross-section of the beam is the sum of the moments over all the elemental
rectangles that constitute the cross-section. This is the bending momentBMof the beam.
)BM=
ÿY
R
(”Ax^2 )
Or,BM=
Y
R
(^1) ÿ
”Ax^2
2
(1.28)
The term(
q
”Ax^2 )=Ig is a characteristic of the geometric shape of the cross-section
of the beam. Therefore,Igis called the geometric moment of inertia of the beam cross-
section.
(For a beam of circular cross-section,Ig=(fir^4 )/ 4 , whereris the beam radius. For
a beam of rectangular cross-section,Ig=(bd^3 )/ 12 , wherebis the breadth anddis the
thickness of the beam.)
Hence
BM=
YIg
R
(1.29)
At equilibrium, the moment of the external bending forces is balanced by the beam’s
bending moment (i.e., the moment of the internal restoring couple).
PH8151 29 LICET